Question

The $ \vartriangle PQR $ is inscribed in a semicircle. $ PQ = PR = 7cm $ .
What is the measure of $ \angle QPR $ ?
Find the area of the shaded region.

Answer 1

Hint: A semicircle is the half of a complete circle, the measure of its full arc is $ 180^\circ \,or\,\pi $ radians. In the given figure, the diameter is also a side of the triangle. By finding out the required angle, we can find the radius of the semicircle and thus the area of the shaded region by subtracting the area of the triangle from the area of the semicircle.

Complete step-by-step answer:
The measure of the angle prescribed by the diameter of the semicircle on its circumference is $ 90^\circ $ . So $ \angle QPR = 90^\circ $
Therefore, $ \vartriangle PQR $ is a right-angled triangle.
Area of a right-angled triangle is,
 $ {A_\vartriangle } = \dfrac{1}{2} \times base \times height = \dfrac{1}{2} \times 7 \times 7 = \dfrac{{49}}{2}\;c{m^2} $
To find the area of the semicircle, we have to find out its radius first.
As $ \vartriangle PQR $ is a right-angled triangle, so by applying Pythagoras theorem, we get –
 $
  Q{R^2} = Q{P^2} + P{R^2} \\
 \Rightarrow Q{R^2} = {7^2} + {7^2} \\
 \Rightarrow Q{R^2} = 2{(7)^2} \\
   \Rightarrow QR = 7\sqrt 2\; cm \;
  $
QR is the diameter of the semicircle, we know that radius is half of the diameter of the circle, so $ r = \dfrac{{7\sqrt 2 }}{2}cm $
The area of the semicircle is $ {A_{semi}} = \dfrac{{\pi {r^2}}}{2} = \dfrac{1}{2} \times \dfrac{{22}}{7} \times {(\dfrac{{7\sqrt 2 }}{2})^2} = \dfrac{{77}}{2}c{m^2} $
The area of the shaded region is $ A = {A_{semi}} - {A_\vartriangle } = \dfrac{{77}}{2} - \dfrac{{49}}{2} = \dfrac{{28}}{2} = 14\;c{m^2} $
Hence the area of the shaded region is $ 14\;c{m^2} $ .
So, the correct answer is “ $ 14\;c{m^2} $ ”.

Additional information:
The sum of all the angles of a triangle is $ 180^\circ $ , so
 $ \angle Q + \angle P + \angle R = 180^\circ $
Equal sides have equal angles opposite to them and vise-versa. As the side $ PQ = PR = 7cm $ so $ \angle Q = \angle R $
Using this result in the obtained equation, we get –
 $
  \angle Q + 90^\circ + \angle Q = 180^\circ \\
 \Rightarrow 2\angle Q = 90^\circ \\
   \Rightarrow \angle Q = \angle R = 45^\circ \;
  $

Note: Along with half of the circumference of the circle, the semicircle also includes the diameter of the circle. We know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle. The angle subtended by the arc QR at centre is $ 180^\circ $ , that’s why the angle subtended by diameter on the circumference of the semicircle is $ 90^\circ $ .